什么是背后的基本原理（X％64）==（X放大器; 63）？ [英] What is the rationale behind (x % 64) == (x & 63)?

问题描述

可能重复：
  位运算和代替模运算

有人能解释，使双方的前pressions当量的理由是什么？我知道这只能是因为64是二的幂，但我怎么能在逻辑上或数学上的分工去按位和？

Can someone explain the rationale that makes both expressions equivalents? I know it only works because 64 is a power of two, but how can I logically or mathematically go from division to bitwise and?

推荐答案

操作 X％64 返回余数时， X 是64，这（假定X> 0）必须是一个介于0和63。让我们来看看这二分：

The operation x % 64 returns the remainder when x is divided by 64, which (assuming x>0) must be a number between 0 and 63. Let's look at this in binary:

63 <子>月 = 0011 1111 <子> B
64 <子>月 = 0100 0000 <子> B

63_dec = 0011 1111_b
64_dec = 0100 0000_b

您可以看到，64的倍数的二进制再presentation必须有6个零结束。因此，当除以64任何数目的余数是原始数，与所有的除了最右边的6那些除去位

You can see that the binary representation of any multiple of 64 must end with 6 zeroes. So the remainder when dividing any number by 64 is the original number, with all of the bits removed except for the 6 rightmost ones.

如果你走位与一个数字63，结果是完全相同的6位。

If you take the bitwise AND of a number with 63, the result is exactly those 6 bits.

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